For a short exact sequence $0 \to G \to H \to K \to 0$ of (discrete) groups with $K$ finite we have, as a consequence of the Hochschild-Serre spectral sequence, that $H^{\ast}(H;\mathbb Q) = H^{\ast}(G;\mathbb Q)^K$. This can be used to see that free and free abelian groups embedd with finite index into groups with trivial rational cohomology: $$H^{\ast}(F_n \rtimes_{\text{sign}} \mathbb Z/2;\mathbb Q) = H^{\ast}(\bigvee_n S^1;\mathbb Q)^{\mathbb Z/2} = H^{\ast}(\text{pt};\mathbb Q).$$ $$H^{\ast}(\mathbb Z^n \rtimes_{\text{sign}} (\mathbb Z/2)^n;\mathbb Q) = H^{\ast}(\prod_n S^1;\mathbb Q)^{(\mathbb Z/2)^n} = H^{\ast}(\text{pt};\mathbb Q).$$ Does this work for every group? Or writing down the opposite:

Is there a group $G$ such that every group $H$ which contains $G$ with finite index has nontrivial rational cohomology?


2 Answers 2


Thompson's group $T$ gives an example, i.e. if $T \le H$ has finite index, then $H^\ast(H;\mathbb{Q}) \neq 0$.

More specifically, there is always a non-trivial class in $H^4(H;\mathbb{Q})$.

Proof: Step 1: $T$ is normal in $H$

Since $T$ has finite index in $H$, $T_0 := \bigcap_{h \in H/T}hTh^{-1}\le T$ is a finite index, normal subgroup of $H$ (and $T$). But $T$ is infinite simple, so $T=T_0$ is normal in $H$.

Step 2: Since $T$ is normal, $H^\ast(H;\mathbb{Q})=H^\ast(T;\mathbb{Q})^H$.

By work of Ghys & Sergiescu, $H^2(T;\mathbb{Z})= \mathbb{Z}^2$ is generated by the Euler class $x'$ of $T$ and the Godbillion-Vey class $y'$. The corresponding rational classes $x=i(x')$ and $y=i(y')$ where $i: H^\ast(T;\mathbb{Z}) \to H^\ast(T;\mathbb{Q})$ generate the rational cohomology ring as $H^\ast(T;\mathbb{Q})=\mathbb{Q}[x,y]/(xy),\,\,|x|=|y|=2$.

I'll show that $x^2 + y^2$ is invariant under the action of $H$. Let $c: T \to T$ be conjugation by an element from $H$. Hence $c^\ast$ induces an isomorphism on $H^2(T;\mathbb{Z})$. Write $c^\ast(x')=ax'+by',\, c^\ast(y')=cx' + dy'$ with integers $a,b,c,d$ s.t. $ad-bc= \pm 1$. Since $c^\ast$ commutes with $i$, we find $c^\ast(x)c^\ast(y)=acx^2 + bdy^2$. But also $c^\ast(x)c^\ast(y)=c^\ast(xy)=0$ because $xy=0$. Hence $ac=0$ and $bd=0$. Thus $$c^\ast(x)=\pm x,\, c^\ast(y)=\pm y\quad \text{or}\quad c^\ast(x)=\pm y\,,c^\ast(y)=\pm x$$ In either case, $c^\ast(x^2+y^2)=x^2+ y^2$. qed


Not a complete answer. I have no idea how to approach this question if $G$ is not required to be normal. If it is, suppose that $G$ has both trivial center and trivial outer automorphism group. (Such groups are called complete.) Then every short exact sequence $1 \to G \to H \to K \to 1$ is trivial in the sense that $H \cong G \times K$, hence if $K$ is finite we have $H^{\bullet}(BH, \mathbb{Q}) \cong H^{\bullet}(BG, \mathbb{Q})$. So it suffices to find $G$ which

  • has trivial center,
  • has trivial outer automorphism group, and
  • has nontrivial rational cohomology.

Such groups ought to exist. Explicitly, I think $\text{Aut}(F_2)$ is an example; apparently it's a result due to Dyer and Formanek that this group is complete, and most likely it has nontrivial rational cohomology...

  • 3
    $\begingroup$ Oh, hmm. If you additionally require that $G$ itself has no finite index subgroups, then it must be normal in any $H$ in which it's finite index. So if anyone can find $G$ with this additional property then that's a counterexample. Maybe the automorphism group of an infinite set? $\endgroup$ Oct 20, 2015 at 7:54
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    $\begingroup$ This is a good strategy. There's every reason to think that such groups should exist. (Groups with no finite-index subgroups are plentiful, and the other properties are sufficiently generic that they should be possible to ensure.) Finitely presented candidates include Wise, Bhattacharjee and Burger--Mozes' examples of groups with no proper finite quotients, and Thompson's infinite simple groups U and V. I think all of these have trivial centre, but the questions of computing the cohomology and outer automorphism groups might require some work. $\endgroup$
    – HJRW
    Oct 20, 2015 at 10:37
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    $\begingroup$ There seems to be a 1996 paper of Brin showing that Out(T) has order 2. So it's not quite what's needed, but very close; I wonder if your argument can be adapted slightly to handle this, using Brin's description of Out(T)? $\endgroup$
    – HJRW
    Oct 20, 2015 at 11:03
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    $\begingroup$ Just a remark, $\mathrm{Aut}(F_2)$ has no rational homology. The first time $\mathrm{Aut}(F_n)$ has rational homology is $n=4$, when $H_4(\mathrm{Aut}(F_4);\mathbb Q)=\mathbb Q$. $\endgroup$
    – Jim Conant
    Oct 21, 2015 at 17:10
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    $\begingroup$ Completeness is unnecessary anyway. As Todd Leason's answer suggests, if I'm reading it correctly, you can replace this condition with the condition that the automorphism group of the rational cohomology has a nontrivial fixed point. (Still need no finite-index subgroups for this approach though.) $\endgroup$ Oct 22, 2015 at 18:22

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